A refrigerator freezes 5 kg of water at `0^(@)C` into ice at `0^(@)C` in a time internal of 20 minutes. Assume that the room temperature is `20^(@)C`. Calculate the minimum power needed to accomplish it.

Answer

`102.5W`

Explanation

Amount of heat required to be taken out
`Q_(2)= 5xx336xx10^(3)J` ,
`T_(2)= 0^(@)C = (0+273)K=273K`
`T_(1)= 20^(@)C= (20+273)K= 293K`
coeff. Of performance of refrigerator
`Q_(2)/W= (T_(2))/(T_(1)-T_(2))=(273)/(293-273)`
`:. W=(Q_(2)xx20)/(273)=(5xx336xx10^(3)xx20)/(273)`
Minimum power needed ,
`P=W/t=(5xx336xx10^(3)xx20)/(273xx20xx60)= 102.5 "watt"`